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== == Cos2x - 1 = [1 - 2sin2 x] - 1 = - 2sin2 x; so [Cos2x - 1] / x = -2 [sinx] [sinx / x] As x approaches 0, sinx / x app 1 while 2 sinx app 0; hence the limit is 0.

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14y ago
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Q: Limit x approaches 0 sinx divide x equals 1 evaluate limit x approaches 0 cos2x-1 divide x?
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